/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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NO

After renaming modulo { a->0, b->1, c->2 }, 
it remains to prove termination of the 3-rule system
{ 0 -> 1 ,
  0 2 1 -> 2 0 1 0 2 ,
  1 2 -> }


Loop of length 19 starting with a string of length 10
using right expansion and the encoding { 0->a, 1->b, ... }:

.acb.bbbccbb
	rule acb-> cabac at position 0
.cabac.bbbccbb
	rule acb-> cabac at position 3
.cabcabac.bbccbb
	rule acb-> cabac at position 6
.cabcabcabac.bccbb
	rule bc->  at position 5
.cabcaabac.bccbb
	rule acb-> cabac at position 7
.cabcaabcabac.ccbb
	rule bc->  at position 6
.cabcaaabac.ccbb
	rule a-> b at position 8
.cabcaaabbc.ccbb
	rule bc->  at position 8
.cabcaaab.ccbb
	rule bc->  at position 7
.cabcaaa.cbb
	rule acb-> cabac at position 6
.cabcaacabac.b
	rule a-> b at position 7
.cabcaacbbac.b
	rule acb-> cabac at position 5
.cabcacabacbac.b
	rule a-> b at position 6
.cabcacbbacbac.b
	rule acb-> cabac at position 8
.cabcacbbcabacac.b
	rule bc->  at position 7
.cabcacbabacac.b
	rule a-> b at position 7
.cabcacbbbacac.b
	rule a-> b at position 9
.cabcacbbbbcac.b
	rule acb-> cabac at position 11
.cabcacbbbbccabac.
	rule a-> b at position 12
.cabcacbbbbccbbac.